Page 308 - Aerodynamics for Engineering Students

290 Aerodynamics for Engineering Students

Now
c, = 1 - (y

(3’
-0.1068 = 1 - -

Hence
q = ~‘(1 - C,)’/’ = 69.7(1.1068)’/2
= 73.2 m s-l
(iii) The total pressure is equal to stream static pressure plus the dynamic pressure and, therefore,
pressure difference corresponding to the reading of the third tube is (PO + fp4) -PO, i.e. is equal
to f p4. Therefore, if the reading is 13
+p4=mr3sini?

2980 = 1000 x 0.85 x 9.807 x 13 x f
whence
13 = 0.712m
Since the total head is greater than the stream static pressure and, therefore, greater than
atmospheric pressure, the liquid in the third tube will be depressed below the zero level, i.e. the
reading will be -0.712m.

Example 6.6 An aircraft is flying at 6100m, where the pressure, temperature and relative
density are 46500Nm-’, -24.6”C, and 0.533 respectively. The wing is vented so that its
internal pressure is uniform and equal to the ambient pressure. On the upper surface of the
wing is an inspection panel 15Omm square. Calculate the load tending to lift the inspection
panel and the air speed over the panel under the following conditions:
(i) Mach number = 0.2, mean C, over panel = -0.8
(ii) Mach number = 0.85, mean C, over panel = -0.5.
(i) Since the Mach number of 0.2 is small, it is a fair assumption that, although the speed over
the panel will be higher than the flight speed, it will still be small enough for compressibility to
be ignored. Then, using the definition of coefficient of pressure (see Section 1.5.3)
c - P1-P
PI --
0.7pMz

pi -p = 0.7pM2 C,, = 0.7 x 46 500 x (0.2)’ x (-0.8)
= -1041 Nm-’

The load on the panel = pressure difference x area
= 1041 x (0.15)2
= 23.4N
Also
e,, = 1 - (y